Tangent and Normal of space curves

Question

I want to know a geometric and an intuitive idea why $\gamma'$ gives tangent and why $\gamma''$ is parallel to the direction of normal for a curve $\gamma: I\rightarrow \mathbb R^3 $ ?
Parametrisation isn't giving me any visualisation.

Answer 1

Let $\gamma:(a,b) \to \mathbb{R}^n$, $s'=\lVert\gamma'\rVert $ (obviously this can be solved to give the arc-length parameter $s$, but we don't need it.

Differentiating once, $$ \gamma' = \frac{\gamma'}{s'} s' = \frac{\gamma'}{\lVert \gamma' \rVert}s' = T s'. $$ The fraction is a unit vector, the tangent vector. This is tangent to the curve for exactly the same reason that the derivative in one dimension gives the tangent: the line $\{\gamma(a)+\gamma'(a)t\mid t \in \mathbb{R}\}$ is the tangent line to the curve. For a curve $\gamma(x)=(x,y)=(x,f(x))$ in the plane, $\gamma'(x)=(1,f'(x))$ gives tangent line $\{(a+t,f(a)+tf'(a))\mid t \in \mathbb{R}\}$; eliminating $t$ gives the ordinary tangent line $Y=f(a)+f'(a)(X-a)$. Hence the tangent vector gives a vector in the direction of the tangent line in this case, and it should be clear that this will work for more complicated curves in more dimensions (write it out in components to make sure).

Now, the normal. Differentiating, $$ \gamma'' = T's' + Ts''. $$ This has a simple geometric interpretation: $ \gamma'' \cdot T = s'' = \lVert \gamma' \rVert' $, so the tangential component tells us about how the parameter deviates from (a multiple of) arc-length.

Thus the normal component is $\gamma''-s''T$. If $s''=0$, therefore, $\gamma'' = T' s'$ is normal to the tangent vector, and so the normal is proportional to $T'$.

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From this we can obtain an expression for the curvature, which is defined by $\dot{T}=\kappa N$ with $\kappa\geq 0$: since $T' = (dT/ds) (ds/dt) = \dot{T} s'$, we have $$ \gamma'' = \kappa N s'^2 + Ts''. $$ The easiest way to go now is to pretend to be in $3$ dimensions briefly, and take the cross product with $\gamma'$, and then the norm: $$ \lVert \gamma' \times \gamma'' \rVert = \kappa \lVert \gamma' \times N \rVert^2 s'^2 = \kappa s'^3, $$ and so $$ \kappa= \frac{\sqrt{\lVert\gamma'\rVert^2\lVert\gamma''\rVert^2-(\gamma'\cdot \gamma'')^2}}{\lVert \gamma' \rVert^3}, $$ which is actually valid in any number of dimensions.